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What percentage of adult women can fit through a door without trouble?Women's heights...

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surfer3 | (Level 1) Valedictorian

Posted May 27, 2012 at 8:02 PM via web

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What percentage of adult women can fit through a door without trouble?

Women's heights are normally distributed with the mean of 63.6 in. and standard deviation of 2.5 in.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 28, 2012 at 4:41 AM (Answer #1)

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You have given that the women's heights are normally distributed with the mean of 63.6 in. and standard deviation of 2.5 in. and the  percentage of adult women can fit through a door without trouble.

This could be done if the height of the door is known but that is not given. I do not see how an assumption can be made that the door is 80 inches high.

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted May 28, 2012 at 3:12 AM (Answer #2)

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Let's start by filling a gap in the problem. The standard door in America has the following dimensions: 36" x 80" (width x height). In other words, in order for a woman to get through a door without trouble, she would need to be maximally 80" tall.

Now, let's move on to the meat of the problem: how can we use the normal distribution to tell us the percentage of women that fit through a door? We are given that the following information:

`mu = 63.6`

`sigma = 2.5`

Here, `mu` is the mean for women's height, and `sigma` is the standard deviation. Using this information, we can take a given height, calculate its "z-score" (how many standard deviations above or below the mean a given value is), and determine the percentage of people fall below that height.

Thus, what we want to do is calculate the z-score for a height of 80" and use that to determine how many women fall below that height. To calculate the z-score, we use the following equation:

`z = (h-mu)/sigma`

Here, `h` is the height that we are testing, which is 80". We can now substitute numbers in for each variable in this equation, allowing us to calculate `z`:

`z = (80 - 63.6)/2.5 = 16.4/2.5 = 6.56`

Thus, we are 6.56 standard deviations above the mean! This z-score is quite high, which we can see by the fact that most z-score tables only go up to 4. Many calculators online simply say the probability of being below this z-score is 1.

At this point, you must use a calculator or computer with a high level of precision to find the probability that a woman's height will be below the z-score of 6.56. Using Matlab, I was able to calculate that the probability of being below the z-score is 0.999968.

In other words, the probability that a woman can fit through a standard door is 0.999968.

Sources:

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted May 28, 2012 at 3:28 AM (Answer #3)

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I apologize. The above answer is for a z-score of 4. The calculation was too close to 1 as to have problems with rounding errors!

The true answer is as follows:

`p(h<=80) = 1-2.6904*10^-11`

In other words, the likelihood that a woman is less than 80" tall is effectively 1. 

However, this probability tells us that the population mean and standard deviation are certainly not true. Based on this number, a woman would never be tall enough to need to lean to get past a door; however, I have certainly seen it before!

I hope this didn't cause you confusion!

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